Let $X$ a normed space over a field $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$ and let $f$ a unbounded linear functional. Let $U \neq \emptyset$ a open set in $X$. I have to prove that $f(U)=\mathbb{K}$.

My attempt. Let $x \in \mathbb{K}$ and since $U \neq \emptyset$ exists $u \in U$ and since $U$ is open exists $\varepsilon>0$ such that $B(u, \varepsilon) \subseteq U$. Since $f$ is unbounded exists $y \in X$ such that $\vert f(x) \lvert > \varepsilon \vert \vert y \vert \vert$. I have noticed that $$f \left( \dfrac{x}{f(y)} y \right)=\dfrac{x}{f(y)} f(y)=x.$$ And also: $$f \left( \dfrac{x}{f(u)} u \right)=\dfrac{x}{f(u)} f(u)=x.$$ I would need to try that $\dfrac{x}{f(y)} y $ or $\dfrac{x}{f(u)} u$ is in $U$. And to do so, prove that is in $B(u, \varepsilon)$. But I'm stuck on it.

  • 1
    $\begingroup$ It suffices to show that the image of the open unit ball $B(0,1)$ is equal $K.$ Indeed any open set contains a ball $B(x_0,r). $ Then $f(B(x_0,r))= f(x_0)+rfB(0,1)).$ By discontinuity the image of the unit ball contains numbers in $K$ of arbitrary large absolute value. Then using $f(\lambda x)=\lambda f(x)$ we can get the conclusion. $\endgroup$ May 14 at 1:55


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